Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.1 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1

Question 1.

Find the equations of tangent and normal to the following curves at the given point on it:

(i) y = 3x^{2} – x + 1 at (1, 3)

Solution:

y = 3x^{2} – x + 1

∴ \(\frac{d y}{d x}=\frac{d}{d x}\) (3x^{2} – x + 1)

= 3 × 2x – 1 + 0

= 6x – 1

∴ \(\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}\) = 6(1) – 1

= 5

= slope of the tangent at (1, 3).

∴ the equation of the tangent at (1, 3) is

y – 3 = 5(x – 1)

∴ y – 3 = 5x – 5

∴ 5x – y – 2 = 0.

The slope of the normal at (1, 3) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}}=-\frac{1}{5}\)

∴ the equation of the normal at (1, 3) is

y – 3 = \(-\frac{1}{5}\)(x – 1)

∴ 5y – 15 = -x + 1

∴ x + 5y – 16 = 0

Hence, the equations of the tangent and normal are 5x – y – 2 = 0 and x + 5y – 16 = 0 respectively.

(ii) 2x^{2} + 3y^{2} = 5 at (1, 1)

Solution:

2x^{2} + 3y^{2} = 5

Differentiating both sides w.r.t. x, we get

= slope of the tangent at (1, 1)

∴ the equation of the tangent at (1, 1) is

y – 1 = \(\frac{-2}{3}\)(x – 1)

∴ 3y – 3 = -2x + 2

∴ 2x + 3y – 5 = 0.

The slope of normal at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,1)}}=\frac{-1}{\left(\frac{-2}{3}\right)}=\frac{3}{2}\)

∴ the equation of the normal at (1, 1) is

y – 1 = \(\frac{3}{2}\)(x – 1)

∴ 2y – 2 = 3x – 3

∴ 3x – 2y – 1 = 0

Hence, the equations of the tangent and normal are 2x + 3y – 5 = 0 and 3x – 2y – 1 = 0 respectively.

(iii) x^{2} + y^{2} + xy = 3 at (1, 1)

Solution:

x^{2} + y^{2} + xy = 3

Differentiating both sides w.r.t. x, we get

= slope of the tangent at (1, 1)

the equation of the tangent at (1, 1) is

y – 1= -1(x – 1)

∴ y – 1 = -x + 1

∴ x + y = 2

The slope of the normal at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{a t(1,1)}}\)

= \(\frac{-1}{-1}\)

= 1

∴ the equation of the normal at (1, 1) is y – 1 = 1(x – 1)

∴ y – 1 = x – 1

∴ x – y = 0

Hence, the equations of tangent and normal are x + y = 2 and x – y = 0 respectively.

Question 2.

Find the equations of the tangent and normal to the curve y = x^{2} + 5 where the tangent is parallel to the line 4x – y + 1 = 0.

Solution:

Let P(x_{1}, y_{1}) be the point on the curve y = x^{2} + 5 where the tangent is parallel to the line 4x – y + 1 = 0.

Differentiating y = x^{2} + 5 w.r.t. x, we get

\(\frac{d y}{d x}=\frac{d}{d x}\)(x^{2} + 5) = 2x + 0 = 2x

\(\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=2 x_{1}\)

= slope of the tangent at (x_{1}, y_{1})

Let m_{1} = 2x_{1}

The slope of the line 4x – y + 1 = 0 is

m_{2} = \(\frac{-4}{-1}\) = 4

Since, the tangent at P(x_{1}, y_{1}) is parallel to the line 4x – y + 1 = 0,

m_{1} = m_{2}

∴ 2x_{1} = 4

∴ x_{1} = 2

Since, (x_{1}, y_{1}) lies on the curve y = x^{2} + 5, y_{1} = \(x_{1}^{2}\) + 5

∴ y_{1} = (2)^{2} + 5 = 9 ……[x_{1} = 2]

∴ the coordinates of the point are (2, 9) and the slope of the tangent = m_{1} = m_{2} = 4.

∴ the equation of the tangent at (2, 9) is

y – 9 = 4(x – 2)

∴ y – 9 = 4x – 8

∴ 4x – y + 1 = 0

Slope of the normal = \(\frac{-1}{m_{1}}=-\frac{1}{4}\)

∴ the equation of the normal at (2, 9) is

y – 9 = \(-\frac{1}{4}\)(x – 2)

∴ 4y – 36 = -x + 2

∴ x + 4y – 38 = 0

Hence, the equations of tangent and normal are 4x – y + 1 = 0 and x + 4y – 38 = 0 respectively.

Question 3.

Find the equations of the tangent and normal to the curve y = 3x^{2} – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.

Solution:

Let P(x_{1}, y_{1}) be the point on the curve y = 3x^{2} – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.

Differentiating y = 3x^{2} – 3x – 5 w.r.t. x, we get

\(\frac{d y}{d x}=\frac{d}{d x}\)(3x^{2} – 3x – 5)

= 3 × 2x – 3 × 1 – 0

= 6x – 3

∴ \(\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=6 x_{1}-3\)

= slope of the tangent at (x_{1}, y_{1})

Let m_{1} = 6x_{1} – 3

The slope of the line 3x – y + 1 = 0

m_{2} = \(\frac{-3}{-1}\) = 3

Since, the tangent at P(x_{1}, y_{1}) is parallel to the line 3x – y + 1 = 0,

m_{1} = m_{2}

∴ 6x_{1} – 3 = 3

∴ 6x_{1} = 6

∴ x_{1} = 1

Since, (x_{1}, y_{1}) lies on the curve y = 3x^{2} – 3x – 5,

\(y_{1}=3 x_{1}{ }^{2}-3 x_{1}-5\), where x_{1} = 1

= 3(1)^{2} – 3(1) – 5

= 3 – 3 – 5

= -5

∴ the coordinates of the point are (1, -5) and the slope of the tangent = m_{1} = m_{2} = 3.

∴ the equation of the tangent at (1, -5) is

y – (-5) = 3(x – 1)

∴ y + 5 = 3x – 3

∴ 3x – y – 8 = 0

Slope of the normal = \(-\frac{1}{m_{1}}=-\frac{1}{3}\)

∴ the equation of the normal at (1, -5) is

y – (-5) = \(-\frac{1}{3}\)(x – 1)

∴ 3y + 15 = -x + 1

∴ x + 3y + 14 = 0

Hence, the equations of tangent and normal are 3x – y – 8 = 0 and x + 3y + 14 = 0 respectively.